If z is a complex number such that z^2+z+1=0, | vedclass

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(A) Given $z^2+z+1=0$,the roots are $z = \omega$ and $z = \omega^2$,where $\omega$ is a complex cube root of unity.For any $n$ not divisible by $3$,$z

Vedclass · Accepted Answer

$4037$

Vedclass · Answer

(A) Given $z^2+z+1=0$,the roots are $z = \omega$ and $z = \omega^2$,where $\omega$ is a complex cube root of unity.For any $n$ not divisible by $3$,$z^n + \frac{1}{z^n} = \omega^n + \omega^{2n} = -1$.For $n$ divisible by $3$,$z^n + \frac{1}{z^n} = \omega^{3k} + \omega^{6k} = 1 + 1 = 2$.There are $2020$ terms in the sum.The number of terms where $n$ is a multiple of $3$ is $\lfloor \frac{2020}{3} floor = 673$.The number of terms where $n$ is not a multiple of $3$ is $2020 - 673 = 1347$.The sum is $1347 	imes (-1)^3 + 673 	imes (2)^3 = -1347 + 673 	imes 8 = -1347 + 5384 = 4037$.

If $z$ is a complex number such that $z^2+z+1=0$,then $\left(z+\frac{1}{z}\right)^3+\left(z^2+\frac{1}{z^2}\right)^3+\left(z^3+\frac{1}{z^3}\right)^3+\ldots+\left(z^{2020}+\frac{1}{z^{2020}}\right)^3=$

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